Question

18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

Answer 1

Answer:

The number of water that must be mixed in the solution is 170.27 mL

Explanation:

given information :

Temperature of water ($T_{1}$) = $23^{o}C$

density of water (ρ) = 1.00 g/mL

Temperature of coffee  ($T_{1}$) = $95^{o}C$

volume of coffee ($V_{2}$) = 180 mL

Mixed Temperature ($T_{mix}$) = $60^{o}C$

to calculate the heat, w use the formula :

Q = m x c x ΔT

where

m = mass of the substance (g)

c = specific heat (J/$g^{o}C$)

ΔT = the temperature change ($^{o}C$)

in the mixture solution, the heat of the water ($Q_{1}$) should be the same as the heat of coffee ($Q_{2}$). Thus,

$Q_{1}$ = $Q_{2}$

$m_{1}$ x $c_{1}$ x Δ$T_{1}$ = $m_{2}$ x $c_{2}$ x Δ$T_{2}$

where

$m_{1}$ is the mass of water

$m_{2}$ is the mass of coffee

$c_{1}$ is the specific heat of water

$c_{2}$ is the specific heat of coffee

Assume coffee and water have the same specific heat. So,

$c_{1}$ = $c_{2}$, we can remove it from the equation.

Hence.

$m_{1}$ x Δ$T_{1}$ = $m_{2}$ x Δ$T_{2}$

we know that

ρ = $\frac{m}{V}$

m = ρ x V, subtitute it to the equation:

ρ$_{1}$ x V$_{1}$ x Δ$T_{1}$ = ρ$_{2}$ x V$_{2}$ x Δ$T_{2}$

$V_{1}$ is the volume of water

coffee and water have the same density, so we can remove the formula

V$_{1}$ x Δ$T_{1}$ = V$_{2}$ x Δ$T_{2}$

V$_{1}$ = (V$_{2}$ x Δ$T_{2}$) / Δ$T_{1}$

V$_{1}$ = V$_{2}$ x ($\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}$

V$_{1}$ = 180 mL x $\frac{(95-60)^{o}C}{(60-23)^{o}C}$

V$_{1}$ = 180 mL x $\frac{(35)^{o}C}{(37)^{o}C}$

V$_{1}$ = 170.27 mL